3.2 Hume-Rothery Compatibility, Interstitial Strain & HSLA Microalloy Scavenging

5. Comparative Metallurgy Tables

Table 3.1-1: Alloying Elements and Interstitial Solutes in Iron

Table 3.1-2: Periodic Properties and Ionization Potentials of Welding Elements

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6. Worked Engineering Calculation: Hume-Rothery Compatibility & Interstitial Strain

Problem Statement

A welding engineer is formulating an experimental filler metal for joining high-strength low-alloy (HSLA) steel. The design requires evaluating:

1. The substitutional solubility limit of molybdenum ($\text{Mo}$) and nickel ($\text{Ni}$) in BCC ferrite ($\alpha\text{-Fe}$) using Hume-Rothery size criteria.
2. The radial misfit strain ($\varepsilon$) generated when interstitial carbon ($r_{\text{C}} = 0.77\ \text{Å}$) occupies an octahedral void in BCC ferrite versus an octahedral void in FCC austenite.

Given data:

• $r_{\text{Fe}} = 1.26\ \text{Å}$
• $r_{\text{Ni}} = 1.25\ \text{Å}$
• $r_{\text{Mo}} = 1.39\ \text{Å}$
• BCC ferrite lattice parameter: $a_{\alpha} = 2.866\ \text{Å}$; octahedral void radius: $r_{\text{oct},\alpha} = 0.19\ \text{Å}$
• FCC austenite lattice parameter: $a_{\gamma} = 3.640\ \text{Å}$; octahedral void radius: $r_{\text{oct},\gamma} = 0.52\ \text{Å}$

Step-by-Step Solution

Step 1: Substitutional Size Misfit Calculation

Calculate the Hume-Rothery size difference parameter $\delta$ for nickel in iron:

$$\delta_{\text{Ni}} = \frac{|r_{\text{Ni}} - r_{\text{Fe}}|}{r_{\text{Fe}}} \times 100\% = \frac{|1.25 - 1.26|}{1.26} \times 100\% = \frac{0.01}{1.26} \times 100\% = 0.79\%$$

Because $0.79\% \ll 15\%$, and nickel has an electronegativity close to iron ($\chi_{\text{Ni}} = 1.91$ vs $\chi_{\text{Fe}} = 1.83$, $\Delta \chi = 0.08$), nickel exhibits extensive substitutional solubility in iron, dissolving up to $10\text{ at}\%$ in BCC ferrite and forming a continuous solid solution in FCC austenite.

Now calculate $\delta$ for molybdenum in iron:

$$\delta_{\text{Mo}} = \frac{|r_{\text{Mo}} - r_{\text{Fe}}|}{r_{\text{Fe}}} \times 100\% = \frac{|1.39 - 1.26|}{1.26} \times 100\% = \frac{0.13}{1.26} \times 100\% = 10.32\%$$

Because $10.32\% < 15\%$, molybdenum satisfies the size factor for substitutional solid solution. However, because $\delta > 8\%$, significant localized lattice strain is introduced around each molybdenum solute atom. This elastic distortion impedes dislocation motion, making molybdenum an effective solid solution strengthener at ambient and elevated temperatures.

Step 2: Interstitial Radial Misfit Strain Calculation

Radial misfit strain is defined as:

$$\varepsilon = \frac{r_{\text{solute}} - r_{\text{void}}}{r_{\text{void}}}$$

For Carbon in an octahedral site of FCC austenite:

$$\varepsilon_{\text{oct},\gamma} = \frac{0.77 - 0.52}{0.52} = \frac{0.25}{0.52} = +0.481\ (+48.1\%\ \text{strain})$$

For Carbon in an octahedral site of BCC ferrite:

$$\varepsilon_{\text{oct},\alpha} = \frac{0.77 - 0.19}{0.19} = \frac{0.58}{0.19} = +3.053\ (+305.3\%\ \text{strain})$$

Engineering Interpretation

The octahedral void in BCC ferrite exhibits a radial misfit strain ($+305.3\%$) more than six times greater than that in FCC austenite ($+48.1\%$). This severe strain explains why:

1. Equilibrium carbon solubility in BCC ferrite at room temperature is negligible ($< 0.005\text{ wt}\%$).
2. Trapping carbon in ferrite during rapid cooling creates the intense tetragonal distortion responsible for the high hardness and brittle behavior of martensite.
3. Carbon atoms in ferrite segregate to the tensile strain fields of edge dislocations, forming Cottrell atmospheres that cause sharp upper and lower yield points and promote strain aging.

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7. Real-World Engineering Application: Interstitial Scavenging in HSLA Steels

In High-Strength Low-Alloy (HSLA) structural steels (such as ASTM A572 Grade 50 or API 5L X70), free interstitial nitrogen and carbon degrade low-temperature impact toughness and cause strain-aging embrittlement in multi-pass welds.

To control this, welding metallurgists employ microalloying with titanium ($0.015\text{--}0.025\text{ wt}\%$) and niobium ($0.02\text{--}0.04\text{ wt}\%$):

• Titanium Nitride Precipitation: Titanium has an exceptionally high affinity for nitrogen. Molten steel droplets precipitate stoichiometric titanium nitride ($\text{TiN}$) at temperatures above $1400^\circ\text{C}$: $$\text{Ti} + \text{N} \rightleftharpoons \text{TiN}(s) \qquad \Delta G^\circ_{1600^\circ\text{C}} \approx -145\ \text{kJ/mol}$$ Because $\text{TiN}$ has high covalent bond energy and a melting point of $2930^\circ\text{C}$, fine $\text{TiN}$ cuboids ($10\text{--}30\text{ nm}$) remain stable at the high temperatures reached in the Coarse-Grained Heat-Affected Zone (CGHAZ).
• Zener Pinning: The stable $\text{TiN}$ particles exert a pinning drag pressure on migrating austenite grain boundaries: $$R_c = \frac{4 r_{\text{precip}}}{3 f_v}$$ where $R_c$ is the limiting austenite grain radius, $r_{\text{precip}}$ is the precipitate radius, and $f_v$ is the volume fraction of precipitates. This pinning limits austenite grain coarsening in the CGHAZ during thermal cycles, promoting the transformation to fine acicular ferrite rather than brittle upper bainite upon cooling.

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8. Common CWEng Exam Traps & Pitfalls

• The BCC Void Volume Fallacy: Candidates often assume that because BCC iron has a lower atomic packing factor ($0.68$) than FCC iron ($0.74$), it must have larger interstitial voids and higher carbon solubility. In reality, BCC's empty volume is dispersed into many small voids; its octahedral void radius ($0.19\ \text{Å}$) is far smaller than FCC's ($0.52\ \text{Å}$). Austenite dissolves nearly $100$ times more carbon than ferrite at their respective eutectoid temperatures.
• Tetrahedral vs. Octahedral BCC Occupancy: In BCC ferrite, the tetrahedral site ($0.36\ \text{Å}$) is larger than the octahedral site ($0.19\ \text{Å}$). However, interstitial carbon and nitrogen occupy the smaller octahedral site because displacing two collinear neighbors along $[001]$ generates less total elastic strain energy than symmetrically displacing four neighbors in tetrahedral coordination.
• Ionization Potential vs. Thermal Conductivity: Candidates frequently conflate an element's first ionization potential with its arc heat generation. Helium has a higher ionization potential ($24.59\text{ eV}$) than argon ($15.76\text{ eV}$), which makes arc initiation harder. However, helium's high thermal conductivity at elevated arc temperatures delivers a broader, deeper penetration profile, not a colder arc.
• Hume-Rothery Size Threshold: The $15\%$ Hume-Rothery rule is a necessary condition for extensive substitutional solubility, not a guarantee. If the electronegativity difference is large (e.g., magnesium in copper), intermetallic precipitation will suppress solubility even if the atomic sizes match within $5\%$.