This is another FITNET fracture case, following the A533B example (§13.2.6, SSTP10). Its focus differs from the A533B example : this time it is a welded stainless-steel wide plate with a through-thickness crack, assessed for ductile tearing (the crack grows stably as the load rises) on the FAD. We work out the assessment point in mechCalc’s BS 7910 Clause 7 fracture assessment calculator, read the chart, and compare point by point with the FITNET literature. The calculator does not provide a ductile tearing analysis, so for each of the three load points we convert the J-R curve to $K_{mat}$ by hand, enter it directly, and get one assessment point per load point.
Source: FITNET FFS Procedure MK7 (2006), Vol. II §13.2.6 (SSTP10 stainless-steel wide plate, through-thickness crack, ductile tearing). The test is a welded wide plate with a through crack under monotonic tension, recording the full ductile-tearing history from initiation (4 MN) to instability (10.83 MN) (inputs and results in Tables 13.11 / 13.12).
🧮 Online calculator: BS 7910 Clause 7 Fracture Assessment Calculator — Try it yourself: the calculation used here combines Annex M (K_I) + Annex P (σ_ref) + the Option 1 FAD, and you can run it online.
1. What this problem assesses
A stainless-steel wide plate, 820 mm wide and 61.2 mm thick, has a through-thickness crack at the weld centre (273 mm total length after fatigue pre-cracking). Under monotonic tension the crack does not snap brittlely; it undergoes ductile tearing: as the load rises, the crack grows a little (Δa) at a time, until instability. FITNET records three key points — initiation (4 MN), mid-tearing (10.35 MN), and instability (10.83 MN).
The object being assessed is whether the plate containing this through crack can keep carrying load. The FAD assessment plots each load point as a coordinate $(L_r,\ K_r)$ and checks whether it lies inside or outside the failure assessment line (FAL).
Figure 1: Geometry and loading of the SSTP10 stainless-steel wide plate. A plate 820 mm wide and 61.2 mm thick, with a transverse through-thickness crack at the weld centre (full length 2a=273 mm, red); monotonic tension is applied at the top and bottom ends, giving a membrane stress σ_m=P/(W·B). The side view on the right shows the crack penetrating the full plate thickness (through-wall) — hence there is no deepest point, and assessment is at a single front. (labels in Chinese)
2. Principle and inputs
Assessment-point coordinates (Option 1):
$$ K_r = \frac{K_I^P + K_I^S}{K_{mat}} + \rho, \qquad L_r = \frac{\sigma_{ref}}{\sigma_Y} $$- Primary stress: monotonic membrane load $\sigma_m = P/(W\cdot B)$ ($W$ width, $B$ thickness).
- $K_I^P$: the stress intensity factor of the through crack (BS 7910 Annex M.3, with finite-width correction), computed by mechCalc.
- $\sigma_{ref}$ / $L_r$: Annex P net-section plastic-collapse reference stress.
- Toughness $K_{mat}$: for ductile tearing the toughness rises with the crack extension $\Delta a$, given by the J-R resistance curve (Table 13.11). The calculator does not provide a ductile tearing analysis, so here $K_{mat}$ is converted by hand for each point with the formula below and entered directly:
| Parameter | Value | Note |
|---|---|---|
| Crack type | through-thickness | weld centre |
| Crack half-length $a$ | 136.5 mm | full length $2a = 273$ mm |
| Plate thickness $B$ / width $W$ | 61.2 / 820 mm | |
| Base material $\sigma_Y$ / $\sigma_U$ | 234 / 594 MPa | stainless-steel base-metal properties |
| Young’s modulus $E$ / $\nu$ | 173 GPa / 0.3 | $E' = 190{,}110$ MPa |
| Primary membrane stress $\sigma_m$ | 79.7 / 206.2 / 215.8 MPa | for 4 / 10.35 / 10.83 MN |
| $K_{mat}$ (from J-R) | 145.9 / 325.2 / 532.9 MPa·m$^{0.5}$ | for $\Delta a$ = 0.2 / 2.38 / 10.94 mm |
Why base-material properties: when FITNET assesses with base-material properties all points lie above the FAL (the conservative, safe side); this is the representative basis for this test, and the one compared against here (FITNET Vol. II §13.2.6, Assessment of Significance of the Results; Fig. 13.32).
3. Try it yourself: entering it in the calculator (using the 4 MN initiation point)
One click: open this example in the calculator . The inputs are filled in for you; the steps below are for checking or entering them by hand.
- Assessment Option (FAD): keep FAD Option “Option 1 (Section 7.3.3)”.
- 1. Crack / Flaw Type: set Geometry to “Flat Plate” and Flaw to “Through-thickness (M.3.1 / P.4.1)”; enter crack depth 136.5 (for a through-thickness crack, $a$ is the half-length, and no half-length $c$ is entered).
- 2. Component Geometry: wall thickness 61.2, component width 820.
- 3. Stress & Polynomial Coefficients: primary membrane stress $P_m=79.71$, primary bending stress $P_b=0$; set “Secondary (residual/thermal) load” to “None (no secondary stress)” (see §5).
- 5. Material: yield strength 234, tensile strength 594, elastic modulus 173.
- 6. Fracture Toughness: enter 145.9 for the fracture toughness (= the J-R curve value at $\Delta a=0.2$ mm; in the first version of the calculator, $K_{mat}$ can only be entered directly).
- Click Run Calculation.
Figure 2: SSTP10 geometry on the calculator’s Schematic tab. A through-thickness crack of full length 2a=273 mm on an 820 mm wide plate; the crack front spans the entire thickness, so a through crack has no ‘deepest point’ and is assessed at a single front.
4. Results
Figure 3: FAD result for the 4 MN initiation point. The assessment point (L_r, K_r)=(0.5107, 0.3844) lies inside the failure assessment line (FAL), in the green acceptable region — verdict ACCEPTABLE. Margin on K_r F_Kr=2.429, reserve factor on load F_load=1.768; K_r ratio 41.2%, L_r ratio 28.9%.
The initiation point (4 MN) assembled by the calculator per Clause 7:
| Quantity | Value | Source |
|---|---|---|
| Primary stress intensity factor $K_I^P$ | 56.08 MPa·m$^{0.5}$ | Annex M.3 through-crack solution |
| Reference stress $\sigma_{ref}$ | 119.5 MPa | Annex P limit load |
| Horizontal $L_r$ | 0.5107 | $\sigma_{ref}/\sigma_Y$ |
| Vertical $K_r$ | 0.3844 | $(K_I^P+0)/145.9 + 0$ |
| Margin on $K_r$, $F_{Kr}$ | 2.429 | $f(L_r)/K_r$, with $f(L_r)=0.9336$ |
| Reserve factor on load $F_{load}$ | 1.768 | factor on the primary load until the point reaches the FAL |
| Verdict | acceptable (large margin) | $K_r < f(L_r)$ and $L_r < L_{r,max}$ |
5. Comparison with FITNET: L_r is close, K_r cannot be compared directly
Putting the three load points (base-material properties) side by side:
| Load (MN) | $\Delta a$ (mm) | $K_{mat}$ (J-R) | mechCalc $L_r$ | FITNET $L_r$ | mechCalc $K_r$ | FITNET $K_r$ |
|---|---|---|---|---|---|---|
| 4.0 (initiation) | 0.2 | 145.9 | 0.511 | 0.51 | 0.384 | 1.15 |
| 10.35 | 2.38 | 325.2 | 1.321 | 1.33 | 0.446 | 0.77 |
| 10.83 (instability) | 10.94 | 532.9 | 1.383 | 1.44 | 0.285 | 0.49 |
The horizontal coordinate $L_r$ is close to FITNET (0.511 vs 0.51, 1.321 vs 1.33, within 0.01 for the first two points; 1.383 vs 1.44 at the instability point, a difference of −4%). $L_r$ is determined only by the primary load and geometry through the limit load, so this column compares the two reference-stress / limit-load solutions for the through crack (FITNET values in Vol. II §13.2.6, Table 13.12).
The vertical coordinate $K_r$ is lower than FITNET because the inputs and the calculation basis differ on the two sides; this is not an error. Two reasons:
- The welding residual stress is not quantified. This plate carries near-yield weld-repair residual stress (tensile at the plate centre; FITNET Vol. II §13.2.6, Fig. 13.29), which raises $K_r$ through $K_I^S$ — it is the main driver pushing the initiation point up to $K_r=1.15$. But the literature gives no value or profile for this residual stress (unlike the A533B example, which explicitly gave $K_I^S=46/5$), so mechCalc cannot feed it in; this post computes only the primary (membrane) side. A rough estimate is that $K_I^S \gtrsim 110$ MPa·m$^{0.5}$ would be needed to reach 1.15.
- Ductile-tearing $K_r$ uses the J route. Along the tearing locus FITNET’s $K_r$ couples the J-R curve with residual stress, which differs from the simplified “primary-side $K_I^P$ ÷ J-R toughness” basis used here.
Note: when the same quantity differs between two procedures, first separate a difference in solution from a missing input. Here the inputs for $L_r$ are complete on both sides, and the difference (at most 4%) comes only from the two limit-load solutions; $K_r$ includes residual stress on one side and not on the other, so the inputs differ, and only $L_r$ is compared.
6. The boundary of this example
The A533B example could match the residual $K_I^S$ point by point because the literature gave measured residual values; for SSTP10 the literature does not quantify the residual stress, so the comparison stops at $L_r$. Compare firmly what can be compared, and state plainly what input is missing: the $L_r$ line (reference stress / limit load) is close on both sides (at most 4% apart), and the gap in $K_r$ can be explained by the missing residual-stress input, not by a difference in calculation method.
🧮 Online calculator: BS 7910 Clause 7 Fracture Assessment Calculator — Try it yourself, the initiation point: Option 1, through crack a=136.5, B=61.2, W=820, P_m=79.71, base material σ_Y=234/σ_U=594/E=173, K_mat=145.9, to obtain L_r=0.511, K_r=0.384, ACCEPTABLE.
Verification manual for this calculator: BS 7910 Fracture Assessment Calculator verification manual