In the table, $F_{Kr}=f(L_r)/K_r$ is the margin on $K_r$: at the same $L_r$, the height of the failure assessment line divided by the $K_r$ of the assessment point. $F_{load}$ is the reserve factor on load: the factor by which all primary loads can be scaled together until the assessment point reaches the failure assessment line, with secondary stresses kept unchanged (it matches the load factor $F^L$ in BS 7910:2019+C1:2025, N.2.4.6). For both, a value above 1 is safe, and 1 means the point is on the line. For examples that give a required toughness, a remaining strength or a limiting depth, the source reports the value at which the point lies on the line, so the example value is 1 by definition.
More background in the engineering series: Residual Stress in A533B Welded Plates ; FITNET Case Studies in FAD Assessment .
The table lists the example value and the MechCalc value for each example. “Agrees” means MechCalc matches the example value within tolerance; “Method difference (see note N)” means the two use different methods or conventions, with the reasons numbered after the table. “[Source N]” points to Source N in the list below the table; it is not repeated when the previous result of the same example has the same source.
| Example | Result | Quantity | Example value | MechCalc | Error (%) |
|---|---|---|---|---|---|
| ASM Handbook Ex. 1: as-welded, Level 2c | Agrees [Source 1] | Margin on $K_r$, $F_{Kr}$ | 1 | 0.9994 | -0.06 |
| Agrees | Reserve Factor on Load $F_{load}$ | 1 | 0.999 | -0.10 | |
| ASM Handbook Ex. 1: PWHT, Level 2c | Agrees [Source 2] | Margin on $K_r$, $F_{Kr}$ | 1 | 1.0002 | +0.02 |
| Agrees | Reserve Factor on Load $F_{load}$ | 1 | 1.0002 | +0.02 | |
| ASM Handbook Ex. 1: as-welded, Level 2b | Method difference (see note 1) [Source 3] | Margin on $K_r$, $F_{Kr}$ | 1 | 0.8962 | -10.38 |
| Method difference | Reserve Factor on Load $F_{load}$ | 1 | 0.8959 | -10.41 | |
| Xie: Q345, a = 8 mm, L = 10 mm | Method difference (see note 2) [Source 4] | Reserve Factor on Load $F_{load}$ | 1 | 0.955 | -4.50 |
| Xie: Q235, a = 10 mm, L = 10 mm | Method difference (see note 3) [Source 5] | Reserve Factor on Load $F_{load}$ | 1 | 0.9473 | -5.27 |
| Xie: Q345, a = 2 mm, L = 40 mm | Method difference (see note 4) [Source 4] | Reserve Factor on Load $F_{load}$ | 1 | 0.9572 | -4.28 |
| Yang 2009 Case 1: a/c = 0.2 | Method difference (see note 5) [Source 6] | Reserve Factor on Load $F_{load}$ | 1 | 1.0618 | +6.18 |
Notes:
- Note 1 (ASM Handbook Ex. 1: as-welded, Level 2b): The handbook’s Level 2b curve and its CTOD conversion (plane strain, X = 2) differ from BS 7910:2019 Option 1 and Eq. 7.16 and 7.17. With both replaced by the handbook’s choices, the result is 1.0006.
- Note 2 (Xie: Q345, a = 8 mm, L = 10 mm): The paper uses the BS 7910:2005 Level 2A line. MechCalc uses Option 1 of the 2019 edition, which lies below the old 2A line at this point, so $F_{load}$ is below 1. Solving Option 1 separately gives a remaining strength of 112.7 MPa against 118 MPa in the paper (ratio 0.955), which matches MechCalc.
- Note 3 (Xie: Q235, a = 10 mm, L = 10 mm): The paper uses the BS 7910:2005 Level 2A line. MechCalc uses Option 1 of the 2019 edition, which lies below the old 2A line at this point, so $F_{load}$ is below 1. Solving Option 1 separately gives a remaining strength of 62.5 MPa against 66 MPa in the paper (ratio 0.947), which matches MechCalc.
- Note 4 (Xie: Q345, a = 2 mm, L = 40 mm): The paper uses the BS 7910:2005 Level 2A line. MechCalc uses Option 1 of the 2019 edition. Here $L_r=1.118>1$, where the two lines differ most, so $F_{load}$ is below 1. Solving Option 1 separately gives a remaining strength of 272.8 MPa against 285 MPa in the paper (ratio 0.957), which matches MechCalc.
- Note 5 (Yang 2009 Case 1: a/c = 0.2): The paper uses the BS 7910:2005 Level 2A line and a $\sigma_{ref}$ that includes a factor of 1.2. MechCalc uses Option 1 of the 2019 edition, and $\sigma_{ref}$ from Eq. P.17 has no 1.2. Both differences come from the edition.
Sources:
- Source 1: ASM Handbook Vol. 11 (2002), H.S. Reemsnyder, Failure Assessment Diagrams, Example 1, Eq. 25–26, 33–41, 45–46, Table 4
- Source 2: ASM Handbook Vol. 11 (2002), H.S. Reemsnyder, Failure Assessment Diagrams, Example 1, Eq. 25–26, 34, 36–41, 45–46, Table 4
- Source 3: ASM Handbook Vol. 11 (2002), H.S. Reemsnyder, Failure Assessment Diagrams, Example 1, Eq. 18, 25–26, 33–41, 43–44
- Source 4: Xie Qi et al., 机械工程学报 (Jixie Gongcheng Xuebao), online first 2021-11-23, Eq. (36)–(38), Tables 2 and 4
- Source 5: Xie Qi et al., 机械工程学报 (Jixie Gongcheng Xuebao), online first 2021-11-23, Eq. (36)–(38), Tables 2 and 6
- Source 6: Yang T., Chen X., Fan Z., ASME PVP2009-77609, Tables 8, 9 and 12