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, re-running each independently with mechCalc’s BS 7910 Clause 7 fracture assessment and cross-checking point by point against 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). FITNET is the unified European structural-integrity assessment procedure; its fracture module shares a common lineage with BS 7910:2019 (the R6/SINTAP family), which makes it a good independent literature anchor for a BS 7910 fracture assessment.
🧮 在线计算器：BS 7910 Clause 7 Fracture Assessment Calculator — The Failure Assessment Diagram (FAD) calculation used here: it combines Annex M (K_I) + Annex P (σ_ref) + Option 1 FAD, and lets you re-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:
$$ \sigma_p = \frac{6M}{W t^2}, \qquad M = 0.0975 P $$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, evaluated at the deepest point (where both the primary and secondary $K_I$ exceed those at the surface point, consistent with where fracture started in the test). 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. The measured through-wall residual stress (transverse to the weld, for $x/t \le 0.5$) is fitted as:
$$ \text{As-welded (LLAW, HLAW):}\quad \sigma^* = -108.35 + 3543.6 (x/t) - 9871.5 (x/t)^2 + 2930.3 (x/t)^3 \ \text{MPa} $$$$ \text{PWHT (LLHT, HLHT):}\quad \sigma^* = -28.032 + 609.14 (x/t) - 1590.8 (x/t)^2 \ \text{MPa} $$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.
[!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 M.4.2 solution (finite-plate surface flaw, polynomial stress), 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
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, 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 linearise a non-linear secondary stress).
2.4 How mechCalc assembles it
mechCalc’s BS 7910 Clause 7 fracture assessment assembles the already-verified 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, feeding $K_I^S$ in directly (as-welded 46 / PWHT 5); - 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 (when secondary stress is present you must compute $\rho$; you cannot set it to zero, or $K_r$ becomes non-conservative);
- FAD Option 1 assessment.
Note: both sides take the residual $K_I^S$ from the same set of measured profiles (46 / 5), so what the cross-check 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.
3. The four problems: assessed one by one
Each cross-check focuses on the FAD coordinates $L_r$ and $K_r$. The mechCalc column gives the full set of intermediate quantities; 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 | ≈48 (Sharples & Clayton) / 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% |
Reading the result: what pushes the assessment 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 point sits far outside the FAL, consistent with the fact that the specimen really did fracture; the two implementations of $K_r$ agree almost digit for digit.
📖 Step-by-step calculation in mechCalc (illustrated walkthrough)
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 | ≈83 / 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% |
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.
📖 Step-by-step calculation in mechCalc (illustrated walkthrough)
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 | ≈198 / 46 | — |
| $\rho$ (Annex R) | 0 ($L_r > L_{r,max}$ cut-off) | FITNET simplified procedure | — |
| $\sigma_{ref}$ [MPa] | 937.5 | — (Sattari-Far / ADINA) | — |
| $L_r$ | 1.80 | 1.72 | +4.8% |
| $K_r$ | 3.94 | 3.79 | +3.9% |
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 residual stress is “diluted” by plasticity) — 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 (mechCalc uses the Annex P closed form, the literature uses 3D finite-element analysis under mismatch); the direction is the same, and mechCalc is slightly more conservative.
📖 Step-by-step calculation in mechCalc (illustrated walkthrough)
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 | ≈187 / 5 | — |
| $\rho$ (Annex R) | 0 ($L_r > L_{r,max}$ cut-off) | FITNET simplified procedure | — |
| $\sigma_{ref}$ [MPa] | 886.7 | — (Sattari-Far / ADINA) | — |
| $L_r$ | 1.71 | 1.63 | +4.6% |
| $K_r$ | 0.60 | 0.57 | +5.2% |
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.
📖 Step-by-step calculation in mechCalc (illustrated walkthrough)
4. The four problems together, and conclusions
Laying the four sets of FAD coordinates side by side:
| 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 independent implementations agree closely. 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 is enough for the two calculations to confirm each other’s Clause 7 assembly.
-
The effect of residual stress is reproduced quantitatively (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 $L_r$ has a clear methodological source. For $L_r > 1$, mechCalc uses an analytical Annex P reference stress while the literature computes the limit load by 3D finite-element analysis; together with the non-linearity of the plastic zone, a deviation of about 4–5% is normal and in the same direction (mechCalc slightly more conservative).
-
The conservatism of the assessment is reasonable. All four assessment points fall outside the FAL, consistent with the fact that the specimens really did fracture; the assessment correctly returns “this condition is not acceptable”, with margin to spare. 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. And mechCalc’s BS 7910 Clause 7 calculation independently reproduces the literature results for all four problems, which shows that it assembles correctly at the step most prone to error — residual stress — and is fit for use in engineering assessment.
Reproducible re-run: the mechCalc driver script
.tempTestScript/golden-verify-fitnet-fad/verify_ex1_residual.py(drives the BS 7910 Clause 7 calculation from the literature inputs, with the residual $K_I^S$ fed in directly).References: the FITNET project’s Case Studies for Fracture (A533B-1 Steel Residual Stress Experiments case); BS 7910:2019+A1:2020, 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).