17.4 AWS A5.10 Aluminum Filler Selection & Dilution and Ferrite-Number Calculations

Aluminum Filler Metal Selection: AWS A5.10

Welding aluminum alloys requires careful consumable matching per AWS A5.10 / A5.10M (Bare Aluminum and Aluminum-Alloy Welding Electrodes and Rods). The primary engineering decision in structural aluminum fabrication is choosing between the two workhorse consumables: ER4043 and ER5356.

+-----------------------------------------------------------------------------------------+
|                               ER4043 VS. ER5356 COMPARISON MATRIX                       |
+------------------------------------+--------------------------+-------------------------+
| Characteristic                     | AWS A5.10 ER4043         | AWS A5.10 ER5356         |
+------------------------------------+--------------------------+-------------------------+
| Principal Alloying System          | Al - 5.0% Si             | Al - 5.0% Mg             |
| Solidification Cracking Resistance | Superior (Al-Si eutectic)| Lower (requires strict  |
|                                    | heals freezing fissures) | dilution management)    |
| Weld Pool Fluidity & Wetting       | Exceptional (flows easily| Viscous, fast-freezing, |
|                                    | with bright appearance)  | requires weaving        |
| All-Weld Shear Strength            | Low (~11–13 ksi / 80 MPa)| High (~17–20 ksi / 130  |
|                                    |                          | MPa; 30-50% stronger!)  |
| Ductility & Impact Toughness       | Moderate                 | High                    |
| Post-Weld Anodizing Color Match    | Poor: turns dark gray /  | Excellent: bright silver|
| (on 6061 base metal)               | charcoal-black           | matches base metal      |
| Continuous Service Temperature     | Unrestricted             | STRICTLY RESTRICTED TO  |
|                                    | (up to ~150°C / 300°F)   | T < 65°C (150°F)!       |
+------------------------------------+--------------------------+-------------------------+

The 65°C (150°F) Service Temperature Limit for 5xxx Alloys

CWEng Critical Metallurgy Rule: The Sensitization of 5xxx Alloys Any 5xxx series aluminum alloy or weld deposit containing more than 3.0% Magnesium (such as ER5356 with ~5.0% Mg, ER5183 with ~4.8% Mg, and base metal 5083 with ~4.5% Mg) is metallurgically unstable at sustained temperatures above 65°C (150°F).

                         SENSITIZATION MECHANISM IN AL-MG ALLOYS (> 65°C)

          Solid Solution Alpha Matrix (Al)           Grain Boundary
       +------------------------------------+--------------------------------+
       | High Magnesium in Solution (~5% Mg)|                                |
       |                                    |  Continuous Network of         |
       |       [Mg] ---------->             |  Beta-Phase (Mg2Al3) Film      |
       |                                    |  - Highly Anodic (-1.24 V)     |
       |                                    |  - Rapidly Corrodes            |
       +------------------------------------+--------------------------------+

1. At temperatures exceeding 65°C, excess magnesium precipitates out of the alpha-aluminum solid solution, forming continuous grain boundary films of the intermetallic compound beta-phase (Mg2Al3).
2. In the galvanic series, beta-phase is extremely anodic (-1.24 V vs SCE) compared to the surrounding alpha-aluminum matrix (-0.84 V vs SCE).
3. In the presence of an electrolyte (marine spray, road deicing salts, or industrial condensate) and sustained tensile stress, galvanic dissolution of the beta-phase network causes rapid Stress Corrosion Cracking (SCC) and catastrophic intergranular exfoliation.
4. Mandatory Rule: For aluminum structural applications exposed to temperatures >= 65°C (150°F)—such as truck bulkheads, engine brackets, and chemical heat exchangers—the engineer must specify ER4043 or low-magnesium ER5554 (Al - 2.7% Mg).

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Comprehensive Worked Engineering Example: Dissimilar Metal Weld Dilution & FN Prediction

Problem Statement

A chemical reactor vessel involves a circumferential butt weld joining ASTM A36 carbon steel to AISI 304L austenitic stainless steel. The joint is welded via mechanized GMAW with an AWS A5.9 ER309L solid wire.

From procedure qualification macroetch measurements, the total joint dilution is D_total = 0.26 (26%), consisting of equal 13% dilution from each base metal (D_A36 = 0.13, D_304L = 0.13). The filler metal provides the remaining fraction: f_wire = 1.00 - 0.26 = 0.74.

The measured certified material test report (CMTR) compositions are:

• ASTM A36: 0.18% C, 0.80% Mn, 0.25% Si, 0.05% Cr, 0.05% Ni, 0.008% N, 0.02% Mo, 0.15% Cu, 0.00% Nb.
• AISI 304L: 0.022% C, 1.45% Mn, 0.45% Si, 18.30% Cr, 8.20% Ni, 0.065% N, 0.20% Mo, 0.20% Cu, 0.00% Nb.
• ER309L: 0.018% C, 1.70% Mn, 0.40% Si, 23.50% Cr, 13.40% Ni, 0.045% N, 0.15% Mo, 0.10% Cu, 0.00% Nb.

Calculate:

1. The Chromium Equivalent (Cr_eq) and Nickel Equivalent (Ni_eq) for each base metal and the filler metal using WRC-1992 formulas.
2. The final diluted weld metal Cr_eq and Ni_eq.
3. The predicted Ferrite Number (FN) and verify that the weld avoids both solidification cracking and martensite.

                               DILUTION LEVER RULE TRIANGLE

                         ER309L Filler Metal (74%)
                                  /\
                                 /  \
                                /    \
                               /  *   \  <-- Weld Metal Puddle Position
                              /        \
                             /__________\
              ASTM A36 (13%)              AISI 304L (13%)

Step-by-Step Solution

Step 1: Compute WRC-1992 Equivalents for Constituents

$$Cr_{eq} = \%Cr + \%Mo + 0.7 \%Nb$$ $$Ni_{eq} = \%Ni + 35 \%C + 20 \%N + 0.25 \%Cu$$

• For ASTM A36 Base Metal: Cr_eq(A36) = 0.05 + 0.02 + 0 = 0.07% Ni_eq(A36) = 0.05 + 35(0.18) + 20(0.008) + 0.25(0.15) = 0.05 + 6.30 + 0.16 + 0.0375 = 6.55%

• For AISI 304L Base Metal: Cr_eq(304L) = 18.30 + 0.20 + 0 = 18.50% Ni_eq(304L) = 8.20 + 35(0.022) + 20(0.065) + 0.25(0.20) = 8.20 + 0.77 + 1.30 + 0.05 = 10.32%

• For ER309L Filler Metal: Cr_eq(wire) = 23.50 + 0.15 + 0 = 23.65% Ni_eq(wire) = 13.40 + 35(0.018) + 20(0.045) + 0.25(0.10) = 13.40 + 0.63 + 0.90 + 0.025 = 14.96%

Step 2: Calculate Diluted Weld Metal Equivalents

$$Cr_{eq(\text{weld})} = 0.13(Cr_{eq(\text{A36})}) + 0.13(Cr_{eq(\text{304L})}) + 0.74(Cr_{eq(\text{wire})})$$

Cr_eq(weld) = 0.13(0.07) + 0.13(18.50) + 0.74(23.65) = 0.0091 + 2.405 + 17.501 = 19.92%

$$Ni_{eq(\text{weld})} = 0.13(Ni_{eq(\text{A36})}) + 0.13(Ni_{eq(\text{304L})}) + 0.74(Ni_{eq(\text{wire})})$$

Ni_eq(weld) = 0.13(6.55) + 0.13(10.32) + 0.74(14.96) = 0.8515 + 1.3416 + 11.0704 = 13.26%

Step 3: Ferrite Number (FN) Prediction & Microstructural Verification

• Ratio calculation: Cr_eq / Ni_eq = 19.92 / 13.26 = 1.502.
• On the WRC-1992 diagram, the solidification mode boundary between AF and FA occurs at Cr_eq / Ni_eq ≈ 1.48. Since 1.502 > 1.48, the weld solidifies via the primary ferrite (FA) mode.
• Evaluating the WRC-1992 iso-ferrite line in this coordinate region yields 6.2 FN.
• Conclusion: The calculated Ferrite Number of 6.2 FN sits safely inside the ideal 3 to 10 FN design window. The weld completely avoids the martensite boundary (which requires Cr_eq < 17% at Ni_eq = 13.26%) and possesses robust resistance to solidification hot cracking.

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Industrial Scenarios & Certified Welding Engineer Exam Traps

Real-World Field Disaster Scenario

A chemical processing facility built an insulated storage tank for hot nitric acid wash solution operating at 80°C (176°F). The structural design specified 5083-H116 aluminum plates joined with ER5356 filler metal to take advantage of 5356's high transverse fillet shear strength. After 14 months of leak-free operation, the entire lower shell-to-bottom fillet joint ruptured catastrophically around 75% of the circumference. Metallographic failure analysis revealed extensive intergranular stress corrosion cracking (SCC) filled with aluminum hydroxide corrosion product. The welding engineer failed to recognize that operating ER5356 (5% Mg) above 65°C caused continuous beta-phase (Mg2Al3) precipitation at grain boundaries, which rapidly dissolved under operating tensile stress in the slightly acidic environment. The entire vessel had to be scrapped and rebuilt using ER4043.

Certified Welding Engineer Exam Traps

Exam Trap 1: ER309L at High Temperatures (> 425°C) Exam questions routinely describe a dissimilar joint between ASTM A387 Gr 11 (Cr-Mo) and 316H stainless in an oil refinery furnace operating at 540°C (1000°F), asking candidates to select the proper filler metal. Candidates instinctively select ER309L. This is wrong! At 540°C, ER309L causes severe carbon migration, forming a brittle decarburized groove in the A387 steel. The correct answer is a nickel-base filler metal: ERNiCr-3 (Inconel 82) or ERNiCrMo-3 (Inconel 625).

Exam Trap 2: Post-Weld Anodizing of 6061 Welded with ER4043 An architect specifies 6061-T6 aluminum handrails with a decorative clear-anodized finish. The shop welds the joints using ER4043 due to its excellent weldability. Upon anodizing, every single weld seam turns an unsightly charcoal-black color, resulting in total rejection. The exam tests your knowledge that ER4043 contains 5% silicon which forms black elemental silicon particles upon anodizing, whereas ER5356 produces an excellent silvery color match.

Exam Trap 3: Ferrite Measurement Units: Percent vs. Ferrite Number Questions test whether candidates know that ferrite in stainless steel is legally measured in Ferrite Number (FN) per AWS A4.2 and ISO 8249, not percent ferrite (% ferrite). While FN roughly approximates volume percentage below 10 FN, magnetic measurement instruments calibrate strictly against NIST primary standards in FN. Expressing ferrite as '%' on a certified welding document is technically non-compliant.