When you run a fitness-for-service (FFS) assessment on a welded pressure component, welding residual stress is almost always one of the hurdles you cannot step around. It is a textbook secondary stress (a self-balancing field): it plays no part in static equilibrium, yet it genuinely raises the driving force at the crack tip. The question engineers have asked for years is this: on the Failure Assessment Diagram (FAD), where does residual stress actually push the assessment point — and how much can post-weld heat treatment (PWHT) pull it back?
The most convincing way to answer that is not a derivation, but a set of large-component tests with real fracture loads and measured residual-stress profiles. That is exactly what this post works through — a set of large welded A533B-1 steel plates tested in four-point bending to fracture. Below we first set out the background and the shared method, then break the four specimens into four problems and assess them one by one, working out each assessment point with mechCalc’s BS 7910 Clause 7 fracture assessment and comparing it with the original literature.
Source: the FITNET project’s Case Studies for Fracture, case “A533B-1 Steel Residual Stress Experiments”; original test report C C France, J K Sharples & C Wignall, AEA Technology Report AEAT-4236 (SINTAP/TASK4/AEAT18, 1998). The analysis details and results are in FITNET FFS Procedure MK7 (2006), Vol. II §13.2.5, Table 13.9 / 13.10, Fig. 13.24. FITNET is the unified European structural-integrity assessment procedure; the Option 1 failure assessment curve used in this case has the same form as Option 1 of BS 7910 Clause 7 (FITNET Vol. II §13.2.5, Table 13.9 step 3; BS 7910:2019, §7.3.3, Eqs. 7.26–7.28), so the case works as an illustration of a BS 7910 fracture assessment.
🧮 Online calculator: BS 7910 Clause 7 Fracture Assessment Calculator — Try it yourself: the Failure Assessment Diagram (FAD) calculation used here combines Annex M (K_I) + Annex P (σ_ref) + Option 1 FAD, and you can run all four problems online.
1. Background: why run a dedicated set of “residual stress” tests
A533B-1 is a typical low-alloy steel for nuclear reactor pressure vessels (RPVs). Once a buried or surface flaw turns up in a weld in this kind of pressure equipment, the assessment must account for welding residual stress as it really is. But residual stress is both hard to measure and changes a great deal with post-weld heat treatment, so for a long time the field lacked a basis calibrated against full-scale components.
To fill that gap, the European SINTAP project of the 1990s (later folded into the FITNET FFS procedure) designed a set of matched tests with a clear aim: drive the FAD from a measured residual-stress profile (not a code envelope curve); quantify the real effect of residual stress through as-welded / PWHT pairs; cover both low and high load-ratio conditions; and check how conservative the assessment is against the real fracture loads.
The tests were built as two pairs, four specimens in all, with names that encode the condition directly — and these are also the subjects of the four problems below:
| Specimen | Weld condition | Test temperature | $L_r$ condition | Fracture load | Fracture behaviour |
|---|---|---|---|---|---|
| LLAW | As-Welded | −120 ℃ | low $L_r$ | 1.27 MN | brittle fracture then arrest (ran about 40 mm in depth, then stopped) |
| LLHT | PWHT | −120 ℃ | low $L_r$ | 2.19 MN | full brittle fracture; load capacity about 1.7× that of LLAW |
| HLAW | As-Welded | −30 ℃ | high $L_r$ | 5.10 MN | clear yielding before fracture + about 5 mm of ductile tearing |
| HLHT | PWHT | −30 ℃ | high $L_r$ | 4.83 MN | clear yielding before fracture + about 4 mm of ductile tearing |
Naming rule: Low / High $L_r$ + As-Welded / Heat-Treated.
Two counter-intuitive effects are already buried in here. In the low-temperature pair, the as-welded LLAW actually broke at a lower load (direct evidence of how harmful residual stress is); in the high-temperature pair, the as-welded HLAW failed at a slightly higher load than the PWHT HLHT. The four problems below pin both of these down on the FAD.
2. Shared method and inputs
2.1 Specimen and loading
Each specimen is a 600 mm × 600 mm A533B-1 steel plate, 70–72 mm thick, with a double-V groove butt weld down the middle and a semi-elliptical surface crack machined into the weld region. The primary load is applied slowly and monotonically by four-point bending, producing a through-wall bending-stress field at the cracked section, converted as (FITNET Vol. II §13.2.5, Table 13.9 step 1):
$$ \sigma_p = \frac{6M}{W t^2}, \qquad M = 0.0975 P $$where $P$ is the fracture load, $M$ is the bending moment, $W = 0.6$ m is the plate width, and $t$ is the plate thickness. The object assessed is the semi-elliptical surface crack in the weld region, assessed at two points, the deepest point and the surface point: both share the same $L_r$, each has its own $K_r$, the flaw is not acceptable if either point lies on or outside the FAL, and each reserve factor takes the smaller of the two points. This follows by analogy the BS 7910:2019+C1:2025 B.5.6 requirement to assess both the deepest point and the surface point (that clause is written for tubular joints). In all four specimens both the primary and secondary $K_I$ are higher at the deepest point than at the surface point, so the verdict is governed by the deepest point in every case. Shared constants: $W = 600$ mm, $E = 210$ GPa, $\nu = 0.3$; every assessment uses the parent-material properties row (the same basis as the literature’s analytical $L_r$ column).
2.2 Measured residual-stress profile → residual K_I^S
The secondary stress comes from welding itself — a through-wall self-balancing residual-stress field, tensile near the surface and compressive in the core. It was not estimated; it was measured layer by layer with the block removal and splitting technique combined with strain gauges (FITNET Vol. II §13.2.5, Figs. 13.21–13.23). The measured through-wall residual stress (transverse to the weld, for $x/t \le 0.5$) is fitted as (FITNET Vol. II §13.2.5, Table 13.9):
$$ \text{As-welded (LLAW, HLAW):}\quad \sigma^* = -108.35 + 3543.6 (x/t) - 9871.5 (x/t)^2 + 2930.3 (x/t)^3 \ \text{MPa} $$$$ \text{PWHT (LLHT, HLHT):}\quad \sigma^* = -28.032 + 609.14 (x/t) - 1590.8 (x/t)^2 \ \text{MPa} $$Integrating the residual profile over the crack front with a weight function gives the deepest-point residual stress intensity factor that enters the FAD: as-welded $K_I^S \approx 46\ \mathrm{MPa\cdot m^{0.5}}$, PWHT $K_I^S \approx 5\ \mathrm{MPa\cdot m^{0.5}}$ — an order-of-magnitude gap, and that is the quantitative reason the as-welded and PWHT assessment points diverge. PWHT lowers the residual stress by at least an order of magnitude relative to the as-welded state, and that is the physical root of the dramatic differences seen later. The surface point is different: both profiles are compressive near the surface, and Annex M.4.2 integration gives a surface-point $K_I$ of $-0.03$ to $-1.5\ \mathrm{MPa\cdot m^{0.5}}$ (worked out once for each specimen’s geometry). When $K_I^S<0$, both $K_I^S$ and $\rho$ are set to 0 (BS 7910:2019, §7.3.6), so the surface-point residual $K_I^S$ is 0 in all four problems.
[!tip] Where does the residual-stress SIF come from?
For a given welding residual-stress profile, you can use mechCalc’s Annex M stress intensity factor calculator, pick the flaw type “M.4.2 · Finite Surface, Polynomial Stress [Fett]”, enter $\sigma_0..\sigma_5$, and integrate out the residual $K_I^S$ that fits your own measured data. We checked this with the as-welded profile from this case: Annex M.4.2 gives $44.84\ \mathrm{MPa\cdot m^{0.5}}$, within 2.5% of the value quoted in FITNET ($46\ \mathrm{MPa\cdot m^{0.5}}$). The procedure, with a term-by-term white-box, is in How is the residual stress intensity factor computed? — integrating a welding residual profile into an SIF with Annex M.4.2 .
2.3 FAD and assessment-point coordinates
All four problems use the BS 7910 Option 1 (standard) Failure Assessment Diagram, with the assessment-point coordinates matched to the Failure Assessment Line (FAL) definition:
$$ K_r = \frac{K_I^P + K_I^S}{K_{mat}} + \rho, \qquad L_r = \frac{\sigma_{ref}}{\sigma_Y} $$$$ f(L_r) = \begin{cases} \left(1 + \tfrac{1}{2}L_r^2\right)^{-1/2}\left[0.3 + 0.7\,e^{-\mu L_r^6}\right], & L_r \le 1 \\ f(L_r{=}1)\,L_r^{(N-1)/(2N)}, & 1 < L_r < L_{r,\max} \\ 0, & L_r \ge L_{r,\max} \end{cases} $$$$ \mu = \min\left(0.001\,E/\sigma_Y,\ 0.6\right), \quad N = 0.3\left(1 - \tfrac{\sigma_Y}{\sigma_U}\right), \quad L_{r,\max} = \frac{\sigma_Y + \sigma_U}{2\sigma_Y} $$Source: BS 7910:2019, §7.3.3, Eqs. 7.26–7.28; $L_{r,\max}$ in §7.3.2, Eq. 7.25; $K_r$ in §7.3.6, Eq. 7.39; $L_r$ in §7.3.7, Eq. 7.40
These three branches happen to split the four problems into two groups: two low-load problems (LLAW, LLHT, $L_r=0.367/0.648$) land in the first branch, where the verdict is fracture-driven; two high-load problems (HLAW, HLHT, $L_r=1.80/1.71$) cross $L_{r,\max}\approx 1.15$ and fall into the third branch.
One key point: residual stress is a secondary stress (BS 7910:2019, §7.3.6 / §7.3.7), so it enters $K_r$ (the vertical axis) only, never $L_r$ (the horizontal axis) — $L_r$ comes solely from the primary stress via the reference stress from Annex P . So the $K_I^S$ from the weight-function integration of the measured profile can simply be fed in directly (which both follows the code’s division of labour and sidesteps the awkward question of how to linearize a non-linear secondary stress).
2.4 How mechCalc assembles it
mechCalc’s BS 7910 Clause 7 fracture assessment assembles Annex M ($K_I$), Annex P ($\sigma_{ref}$), Annex R ($\rho$), and Option 1 FAD following the Clause 7 procedure.
The input mapping is the same for all four problems:
crack_type = surface(semi-elliptical surface crack), with $a_0$, $2c_0$, $B=t$, $W=600$ per problem;- primary stress $P_b = \sigma_p$ (pure bending, $P_m = 0$);
- secondary stress
kis_source = direct, with the directly entered $K_I^S$ in two fields: as-welded 46 / PWHT 5 at the deepest point, and 0 at the surface point (§2.2); - reference stress $\sigma_{ref}$ and $L_r$ from the Annex P semi-elliptical surface-crack limit load (parent-material properties);
- plasticity interaction with $\rho$ computed by Annex R at each of the two points (when secondary stress is present you must compute $\rho$; you cannot set it to zero, or $K_r$ becomes non-conservative; BS 7910:2019, §7.3.6, with $\rho$ from Annex R);
- FAD Option 1 assessment at both the deepest point and the surface point.
Note: both sides take the residual $K_I^S$ from the same set of measured profiles (46 / 5), so what the comparison really tests is each implementation’s own primary-SIF solution, reference-stress solution, and FAD assembly. The primary SIF and limit-load solutions come from different sources (the literature uses Sharples & Clayton / Sattari-Far, mechCalc uses BS 7910 Annex M / Annex P), so they are different solutions for the same physical quantity, and a difference of a few percent is normal. Note: the FITNET original (Vol. II §13.2.5, Table 13.10) lists only $K_r$ and $L_r$, not $K_I^P$, so the FITNET column of $K_I^P$ in the tables below shows “—”.
3. The four problems: assessed one by one
Each comparison focuses on the FAD coordinates $L_r$ and $K_r$. The mechCalc column gives the full set of intermediate quantities at the governing deepest point, with the surface point in an extra row; the FITNET column takes the published parent-material-property row and the analytical limit-load $L_r$.
3.1 Problem 1 — LLAW (as-welded, −120 ℃, low L_r)
Brittle-fracture dominated, with large residual stress. In the test, LLAW broke at just 1.27 MN — the lowest of the four — and then arrested.
Inputs
| Parameter | Value |
|---|---|
| Crack depth $a_0$ / full length $2c_0$ | 19.4 / 175 mm |
| Plate thickness $t = B$ | 71 mm |
| Bending stress $\sigma_p$ (fracture load 1.27 MN) | 245 MPa ($P_m=0$) |
| Parent $\sigma_Y$ / $\sigma_U$ | 618 / 791 MPa |
| Fracture toughness $K_{mat}$ (as-welded weld, −120 ℃) | 37 MPa·m$^{0.5}$ |
| Residual $K_I^S$ (direct) | 46 MPa·m$^{0.5}$ |
Results (FITNET ↔ mechCalc)
| Quantity | mechCalc | FITNET | Difference |
|---|---|---|---|
| $K_I^P$ / $K_I^S$ [MPa·m$^{0.5}$] | 48.13 / 46 | — / 46 | — |
| $\rho$ (Annex R) | 0.046 | FITNET simplified procedure | — |
| $\sigma_{ref}$ [MPa] | 226.5 | — (Sattari-Far) | — |
| $L_r$ | 0.367 | 0.36 | +1.9% |
| $K_r$ | 2.59 | 2.58 | +0.4% |
| Surface point $K_I^P$ / $K_r$ (mechCalc) | 33.86 / 0.915 (inside the FAL) | — | — |
Reading the result: what pushes the deepest point up to $K_r = 2.59$ is mainly the 46 MPa·m$^{0.5}$ residual $K_I^S$ (nearly half of $K_I^P + K_I^S$). The deepest point sits far outside the FAL, consistent with the fact that the specimen really did fracture; the surface point lies inside the FAL, but one unacceptable point is enough to make the flaw not acceptable. The two implementations of $K_r$ agree almost digit for digit.
📖 This problem, step by step (illustrated)
3.2 Problem 2 — LLHT (PWHT, −120 ℃, low L_r)
Same temperature and same region as LLAW; the only variable is that PWHT was applied — the residual stress is relaxed by an order of magnitude. Load capacity rises to 2.19 MN (about 1.7× that of LLAW).
Inputs
| Parameter | Value |
|---|---|
| Crack depth $a_0$ / full length $2c_0$ | 18.6 / 174 mm |
| Plate thickness $t = B$ | 71 mm |
| Bending stress $\sigma_p$ (fracture load 2.19 MN) | 424 MPa ($P_m=0$) |
| Parent $\sigma_Y$ / $\sigma_U$ | 596 / 772 MPa |
| Fracture toughness $K_{mat}$ (PWHT weld, −120 ℃) | 46 MPa·m$^{0.5}$ |
| Residual $K_I^S$ (direct) | 5 MPa·m$^{0.5}$ |
Results (FITNET ↔ mechCalc)
| Quantity | mechCalc | FITNET | Difference |
|---|---|---|---|
| $K_I^P$ / $K_I^S$ [MPa·m$^{0.5}$] | 82.97 / 5 | — / 5 | — |
| $\rho$ (Annex R) | 0.010 | FITNET simplified procedure | — |
| $\sigma_{ref}$ [MPa] | 386.0 | — (Sattari-Far) | — |
| $L_r$ | 0.648 | 0.64 | +1.2% |
| $K_r$ | 1.92 | 1.89 | +1.7% |
| Surface point $K_I^P$ / $K_r$ (mechCalc) | 56.38 / 1.23 (outside the FAL) | — | — |
Reading the result: note one contrast — LLHT’s primary SIF ($K_I^P = 83$) is in fact higher than LLAW’s (48), yet its assessment point $K_r = 1.92$ is lower than LLAW’s 2.59. The whole difference is in the residual $K_I^S$: 46 as-welded, only 5 after PWHT. That turns “residual stress is harmful” from a qualitative statement into a concrete increment on the FAD’s vertical axis.
📖 This problem, step by step (illustrated)
3.3 Problem 3 — HLAW (as-welded, −30 ℃, high L_r)
Warmed up to −30 ℃ and loaded harder into the high-$L_r$ (large-plasticity) region, still as-welded. Failure load 5.10 MN.
Inputs
| Parameter | Value |
|---|---|
| Crack depth $a_0$ / full length $2c_0$ | 19.0 / 174 mm |
| Plate thickness $t = B$ | 70 mm |
| Bending stress $\sigma_p$ (fracture load 5.10 MN, elastic conversion) | 1015 MPa ($P_m=0$) |
| Parent $\sigma_Y$ / $\sigma_U$ | 520 / 677 MPa |
| Fracture toughness $K_{mat}$ (as-welded weld, −30 ℃) | 62 MPa·m$^{0.5}$ |
| Residual $K_I^S$ (direct) | 46 MPa·m$^{0.5}$ |
Results (FITNET ↔ mechCalc)
| Quantity | mechCalc | FITNET | Difference |
|---|---|---|---|
| $K_I^P$ / $K_I^S$ [MPa·m$^{0.5}$] | 198.2 / 46 | — / 46 | — |
| $\rho$ (Annex R) | 0 ($L_r \ge 1.05$, Annex R, Eq. R.2) | FITNET simplified procedure | — |
| $\sigma_{ref}$ [MPa] | 937.5 | — (Sattari-Far) | — |
| $L_r$ | 1.80 | 1.72 | +4.8% |
| $K_r$ | 3.94 | 3.79 | +3.9% |
| Surface point $K_I^P$ / $K_r$ (mechCalc) | 137.99 / 2.23 (also past the cut-off) | — | — |
Reading the result: $L_r = 1.80 > L_{r,max}\approx1.15$, so the verdict is plastic collapse. $K_r$ reaches 3.94, but this time the main driver is not the residual stress (in the high-$L_r$ region plastic deformation weakens the effect of secondary stress, and the Annex R $\rho$ falls to zero for $L_r \ge 1.05$; BS 7910:2019, Annex R, Eq. R.2) — it is the low toughness ($K_{mat} = 62$). The roughly 4–5% deviation at high $L_r$ comes from the difference in the reference-stress solution (both are closed-form, but the limit-load solutions differ: mechCalc uses Annex P, FITNET uses the Sattari-Far surface-crack limit-load formula; FITNET Vol. II §13.2.5, Table 13.9 step 8 and reference [13.5]); the direction is the same, and mechCalc is slightly more conservative.
📖 This problem, step by step (illustrated)
3.4 Problem 4 — HLHT (PWHT, −30 ℃, high L_r)
Same temperature and region as HLAW, but with PWHT applied. The thing to watch is the double benefit of PWHT at −30 ℃: it both lowers the residual stress and brings the toughness back up by an order of magnitude ($K_{mat}$ jumps from 62 to 321).
Inputs
| Parameter | Value |
|---|---|
| Crack depth $a_0$ / full length $2c_0$ | 19.0 / 174 mm |
| Plate thickness $t = B$ | 70 mm |
| Bending stress $\sigma_p$ (fracture load 4.83 MN, elastic conversion) | 960 MPa ($P_m=0$) |
| Parent $\sigma_Y$ / $\sigma_U$ | 520 / 677 MPa |
| Fracture toughness $K_{mat}$ (PWHT weld, −30 ℃, off the lower shelf) | 321 MPa·m$^{0.5}$ |
| Residual $K_I^S$ (direct) | 5 MPa·m$^{0.5}$ |
Results (FITNET ↔ mechCalc)
| Quantity | mechCalc | FITNET | Difference |
|---|---|---|---|
| $K_I^P$ / $K_I^S$ [MPa·m$^{0.5}$] | 187.5 / 5 | — / 5 | — |
| $\rho$ (Annex R) | 0 ($L_r \ge 1.05$, Annex R, Eq. R.2) | FITNET simplified procedure | — |
| $\sigma_{ref}$ [MPa] | 886.7 | — (Sattari-Far) | — |
| $L_r$ | 1.71 | 1.63 | +4.6% |
| $K_r$ | 0.60 | 0.57 | +5.2% |
| Surface point $K_I^P$ / $K_r$ (mechCalc) | 130.51 / 0.41 (also past the cut-off) | — | — |
Reading the result: same plastic-collapse verdict ($L_r > L_{r,max}$), but $K_r$ drops from HLAW’s 3.94 to 0.60 — almost entirely because $K_{mat}$ rises from 62 to 321. In the high-$L_r$ region the governing factors swing back to toughness and net-section yielding.
📖 This problem, step by step (illustrated)
4. The four problems together, and conclusions
Laying the four sets of FAD coordinates side by side (mechCalc values are at the governing deepest point):
| Specimen | mechCalc $L_r$ | FITNET $L_r$ | $\Delta L_r$ | mechCalc $K_r$ | FITNET $K_r$ | $\Delta K_r$ |
|---|---|---|---|---|---|---|
| LLAW | 0.367 | 0.36 | +1.9% | 2.59 | 2.58 | +0.4% |
| LLHT | 0.648 | 0.64 | +1.2% | 1.92 | 1.89 | +1.7% |
| HLAW | 1.80 | 1.72 | +4.8% | 3.94 | 3.79 | +3.9% |
| HLHT | 1.71 | 1.63 | +4.6% | 0.60 | 0.57 | +5.2% |
A few conclusions:
-
The two procedures give closely matching results. At low $L_r$ the $K_r$ values are almost digit for digit (LLAW 2.59 vs 2.58), with the $L_r$ error under 2%; at high $L_r$ the error is still only about 4–5%. Given that the primary-SIF and limit-load solutions come from different sources, this level of agreement shows that both procedures place the four assessment points in the same position and reach the same verdicts.
-
The effect of residual stress becomes a concrete number (Problems 1 and 2). The as-welded LLAW $K_r$ (2.59) is clearly higher than the PWHT LLHT (1.92), even though LLHT actually has the higher primary SIF — what pushes the as-welded point up is the 46 MPa·m$^{0.5}$ residual $K_I^S$.
-
The benefit of PWHT is twofold (Problems 3 and 4 compared): it both relaxes the residual stress ($K_I^S$ 46 → 5) and restores toughness ($K_{mat}$ 62 → 321). The two paths stack, so the assessment point drops markedly in both the low- and high-$L_r$ regions — which is precisely the quantitative justification for insisting on PWHT for important welds in practice.
-
The difference at high $L_r$ comes from two different limit-load solutions. For $L_r > 1$, mechCalc uses the Annex P reference-stress solution, while FITNET uses the Sattari-Far surface-crack limit-load formula (Table 13.9 step 8, reference [13.5]); the two closed-form solutions differ by about 4–5% in the high-$L_r$ region, in the same direction (mechCalc’s $L_r$ is larger, slightly more conservative). FITNET also ran 3D finite-element analysis (ADINA) for the strength-mismatch case (the $L_r$ = 2.36 / 2.25 column), which is outside the comparison here.
-
The conservatism of the assessment is reasonable. The deepest point of all four specimens falls outside the FAL (only the LLAW surface point lies inside the curve, which does not change the verdict), and all four are judged not acceptable, consistent with the fact that the specimens really did fracture; the assessment correctly returns “this condition is not acceptable”, with margin to spare; the FITNET report likewise calls the assessment “considerably conservative” (FITNET Vol. II §13.2.5, Assessment of Significance of the Results). For the safety assessment of a flawed structure, this kind of right-direction result with a conservative margin is exactly what you want.
In one sentence: this set of A533B-1 tests answers “where does residual stress push the FAD assessment point” very concretely — in the brittle-fracture-dominated low-$L_r$ region, as-welded residual stress can contribute a $K_I^S$ on the order of the primary stress, lifting the assessment point markedly; PWHT pulls it back through the combination of relaxing the residual stress and restoring toughness. The four assessment points worked out along the BS 7910 Clause 7 route agree with the literature results, which supports treating residual stress as a secondary stress that enters the vertical axis $K_r$ only, through $K_I^S$, for this series of large-component tests.
References: FITNET FFS Procedure MK7 (2006), Vol. II, §13.2.5 (A533B-1 Steel Residual Stress Experiments case); BS 7910:2019+C1:2025, Clause 7 (§7.3.3 / §7.3.6 / §7.3.7), Annex M / Annex P / Annex R; C C France, J K Sharples & C Wignall, AEAT-4236, SINTAP/TASK4/AEAT18 (1998).
Verification manual for this calculator: BS 7910 Fracture Assessment Calculator verification manual