AS 4100:2020 Fatigue — Verified Benchmark (Hand Calcs vs SDC Verifier)

Test model

Jacket model was used to test the implementation of the code. Structure is modelled mainly with beam elements with the bottom section modeled with shell elements.

The model was constrained on bottom edges of the bottom plate sections. Gravity, loads and wind (only on beam part of the model and not on the bottom structures) were defined:

Several load cases were considered and 2 load groups were created:

Overall – no wind
Overall – wind

Due to the complexity of the model and load cases, in order to find most loaded plate – with the greatest weld stress range (delta stress), FEM analysis was used with following results (From the left: Overall – no wind, Overall – wind, Element with greatest delta stress):

Most loaded shell element (for both load groups) is 23475:

Load Group: Overall – no wind:

∆𝜎_{𝑥_𝑛𝑤} = 38790164 𝑃𝑎

∆𝜎_{𝑦_𝑛𝑤} = 88754472 𝑃𝑎

∆_𝜏𝑛𝑤 = 4202944 𝑃𝑎

Load Group: Overall – wind:

∆𝜎_{𝑥_𝑤} = 47110660 𝑃𝑎

∆𝜎_{𝑦_𝑤} = 107948624 𝑃𝑎

∆𝜏_𝑤 = 4865480 𝑃𝑎

Model material is structured steel with following properties:

Young Modulus: 𝐸 = 200 𝐺𝑃𝑎

Poisson Ratio: 𝜈 = 0.3

Mass Density: 𝜌 = 7850 𝑘𝑔/𝑚3

Yield Stress: 𝑓_𝑦 = 235 𝑀𝑃𝑎

The element is part of the fillet welded shell with thickness: 𝑡_𝑝 = 0.12 𝑚

Detail category for the element was chosen according to the standard (Section 11/Table11.5.1):

For normal directions: 90

For shear directions: 80

Fatigue group consisting of both load groups (Overall – no wind and Overall – wind) wasconsidered with amount of cycles 𝑛_𝑠𝑐 = 250000 for each load group (0.5 million cyclesin total).

Capacity factor was set (Section 11/[11.1.5]): 𝜙 = 0.7

Hand calculations

In order to check the results, analytical calculations were first carried out:

It was checked if standard applies to the model (Section 1/[1.1.2]):

𝑡_𝑝 ≥ 3 𝑚𝑚

120 𝑚𝑚 ≥ 3 𝑚𝑚

thickness requirement met

𝑓_𝑦 ≤ 690 𝑀𝑃𝑎

235 𝑀𝑃𝑎 ≤ 690 𝑀𝑃𝑎

yield stress requirement met

Following hand calculations are performed for the element 23475:

Then limitations were checked (Section 11/[11.1.3]):

Greatest magnitude of stress in all cycles was checked by FEM: 𝑎𝑏𝑠(𝜎_𝑚𝑎𝑥) = 108 𝑀𝑃𝑎

Greatest stress range: 𝑓∗ = 107948624 𝑃𝑎

𝜎_𝑚𝑎𝑥 ≤ 𝑓_𝑦 and 𝑓∗ ≤ 1.5𝑓_𝑦

108 𝑀𝑃𝑎 < 235 𝑀𝑃𝑎 limitation met

108 𝑀𝑃𝑎 < 352.5 𝑀𝑃𝑎 limitation met

As the element is a part of fillet welded plate and plate thickness is greater than 25 𝑚𝑚,thickness correction factor was calculated (Section 11/[11.1.6])

\[ \beta_{tf} = \left( \frac{25\,\text{mm}}{t_p} \right)^{0.25} = 0.676 \]

Exception from assessments were checked (Section 11/[11.4]):

\[ f^{*} < \phi \cdot 27\,\text{MPa}
\quad \text{or} \quad
n_{sc} < 2\times 10^{6}\!\left( \frac{\phi \cdot 36\,\text{MPa}}{f^{*}} \right) \]

Number of cycles: 𝑛_𝑠𝑐 = 500000 and greatest stress range: 𝑓^∗ = 107948624 𝑃𝑎

107948624 𝑃𝑎 > 18900000 𝑃𝑎 Not met

500000 > 466889 Not met

Another exception from assessments were checked (Section 11/[11.7]):

𝑓^∗ < 𝜙𝑓_3𝑐

Greatest stress range for normal directions: 𝑓^∗ = 107948624 𝑃𝑎

For class 90: 𝑓_3𝑐 = 𝛽_𝑡𝑓 ∗ 𝑓₃ = 0.676 ∗ 66 𝑀𝑃𝑎 = 44.6 𝑀𝑃𝑎

107948624 𝑃𝑎 > 32690000 𝑃𝑎 Not met

Exception are not applicable for the element – Fatigue assessment is required (Section11/[11.8.2]):

For normal stresses (with greatest stress range in normal direction taken from both Load Groups: 𝑓^*_𝑛𝑤 = 88754472 𝑃𝑎 and 𝑓^*_𝑤 = 107948624 𝑃𝑎):

\[ n_{sc} \ge \frac{5 \times 10^{6}\,(\phi f_{3c})^{3}}{(f^{*}_{nw})^{3}} \]

\[ n_{sc} \ge \frac{5 \times 10^{6}\,(\phi f_{3c})^{3}}{(f^{*}_{w})^{3}} \]

For shear stresses (with shear stresses from both Load Groups: \( f^{*}_{\tau nw} \) = 4202944 𝑃𝑎 and \( f^{*}_{\ tau w} \)
 = 4865480 𝑃𝑎 ):

\[ n_{sc} \ge \frac{2 \times 10^{6}\,(\phi f_{rsc})^{5}}{(f^{*}_{\tau \mathrm{nw}})^{5}} \]

\[ n_{sc} \ge \frac{2 \times 10^{6}\,(\phi f_{rsc})^{5}}{(f^{*}_{\tau \mathrm{w}})^{5}} \]

For selected categories:

For normal directions: 90 𝑀𝑃𝑎 → 𝑓₃ = 66 𝑀𝑃𝑎 → 𝑓_3𝑐 = 𝛽_𝑡𝑓 ∗ 𝑓₃ = 44.60 𝑀𝑃𝑎

For shear directions: 80 𝑀𝑃𝑎 → 𝑓_𝑟𝑠 = 80 𝑀𝑃𝑎 → 𝑓_𝑟𝑠𝑐 = 𝛽_𝑡𝑓 ∗ 𝑓_𝑟𝑠 = 54.08 𝑀𝑃𝑎

Number of cycles was calculated for:

Normal stress directions:

\[ n_{nw}
= \frac{5 \times 10^{6}\,(\phi f_{3c})^{3}}{(f^{*}_{nw})^{3}}
= 217\,619 \]

\[ n_{w}
= \frac{5 \times 10^{6}\,(\phi f_{3c})^{3}}{(f^{*}_{w})^{3}}
= 120\,953 \]

Shear stress direction:

\[ n_{\tau \mathrm{nw}}
= \frac{2 \times 10^{6}\,(\phi f_{rsc})^{5}}{(f^{*}_{\tau \mathrm{nw}})^{5}}
= 1.1856 \times 10^{11} \]

\[ n_{\tau \mathrm{w}}
= \frac{2 \times 10^{6}\,(\phi f_{rsc})^{5}}{(f^{*}_{\tau \mathrm{w}})^{5}}
= 0.5703 \times 10^{11} \]

Amount of cycles for normal stress directions is smaller than for shear stress directionso fatigue damage is calculated with results for normal stress directions:

𝐹𝑎𝑡𝑖𝑔𝑢𝑒𝐷𝑎𝑚𝑎𝑔𝑒 = \( \frac{n_{sc}}{n} \)

𝐹𝑎𝑡𝑖𝑔𝑢𝑒𝐷𝑎𝑚𝑎𝑔𝑒_𝑛𝑤 = 1.15

𝐹𝑎𝑡𝑖𝑔𝑢𝑒𝐷𝑎𝑚𝑎𝑔𝑒_𝑤 = 2.07

And summed fatigue damage:𝐹𝐷_{𝑆𝑢𝑚𝑚𝑒𝑑} = 𝐹𝑎𝑡𝑖𝑔𝑢𝑒𝐷𝑎𝑚𝑎𝑔𝑒_𝑛𝑤 + 𝐹𝑎𝑡𝑖𝑔𝑢𝑒𝐷𝑎𝑚𝑎𝑔𝑒_𝑤

𝐹𝐷_{𝑆𝑢𝑚𝑚𝑒𝑑} = 3.22

SDC Verifier check results

In SDC Verifier standard was added with same assumption and check was performed:

Result comparison

Results comparison between hand calculations and SDC Verifier check: 

Results from SDC Verifier are the same as those obtained with hand calculations.

This benchmark shows SDC Verifier’s AS 4100 fatigue implementation reproduces manual calculations step-by-step on a realistic jacket model. The governing fillet-welded plate fails with summed damage 3.22, matching the hand-calc route. The check surfaces the critical weld and provides auditor-ready, clause-mapped evidence. (Design levers: reduce stress range, upgrade detail category, increase thickness, or reduce cycles.)